Question

A solid body rotates about a stationary axis so that its angular velocity depends on the rotational angle $ as a = COQ - faj) where C0Q and k are positive constants. At the moment f = 0, (j> = 0, the time dependence of rotation angle is:

Answer 1

Answer on Question #66183-Physics Mechanics-Relativity

A solid body rotates about a stationary axis so that its angular velocity depends on the rotation angle $\phi$ as $\omega = \omega 0 - a\phi$ , where $\omega 0$ and $a$ are positive constants. At the moment $t = 0$ the angle $\phi = 0$ . The time dependence of rotation angle is:

Solution

$$\omega = \omega_0 - a\phi$$ $$\omega = \frac{d\phi}{dt} = \omega_0 - a\phi$$ $$dt = \frac{d\phi}{\omega_0 - a\phi}$$ $$\int dt = \int \frac{d\phi}{\omega_0 - a\phi}$$ $$t + c = -\frac{1}{a}\ln(\omega_0 - a\phi)$$ $$\ln(\omega_0 - a\phi) = -a(t + c)$$ $$\omega_0 - a\phi = C e^{-at}$$ $$\phi = \frac{\omega_0}{a} - C e^{-at}$$ $$\phi(0) = \frac{\omega_0}{a} - C = 0 \rightarrow \frac{\omega_0}{a} = C$$

Thus,

$$\phi = \frac{\omega_0}{a}(1 - e^{-at}).$$

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